Putting the Second Law to Work

Shaun Williams, PhD

Free Energy Functions

• Previously we found that \( \Delta S_{universe}>0 \) for spontaneous processes.
• We need a more useful expression for determining if a system is spontaneous
• We know that: \[ \Delta S_{univ}=\Delta S_{sys}+\Delta S_{surr} \]
• We also know that \[ \Delta S = -\frac{q_{rev}}{T} \]

Building a More Useful Equation

• We also know that at constant pressure \[ \Delta H = q_P \]
• This means that at constant pressure and temperature \[ \Delta S_{surr}=-\frac{\Delta H_{sys}}{T} \]
• So, for a spontaneous process \[ \Delta S_{univ} = \Delta S_{sys} - \frac{\Delta H_{sys}}{T} \ge 0 \]

Almost There!

\[ \Delta S_{sys} - \frac{\Delta H_{sys}}{T} \ge 0 \]

• Multiplying both sides by \(-T\) \[ \Delta H - T\Delta S\le 0 \]
• Note: At constant volume and pressure \[ \Delta U - T\Delta S \le 0 \]
• We can handle non-constant temperatures by looking at infinitesimal changes. \[ dH-TdS\le 0 \text{ and } dU-TdS\le 0 \]

Gibbs Function

\[ \Delta H - T\Delta S\le 0 \]

• This equation suggests a new thermodynamic function called the Gibbs' Function (or Gibbs' Free Energy) \[ G\equiv H - TS \]
• A change in this function is given by \[ \Delta G = \Delta H - \Delta \left( TS \right) \]
• At constant temperature \[ \Delta G = \Delta H - T \Delta S \]
• For a spontaneous process, \(\Delta G < 0 \)

Spontaneous Reactions

\[ \Delta G = \Delta H - T \Delta S \lt 0 \]

Consider the factors driving the following spontaneous reactions: \[ \chem{NaOH(s)\rightarrow Na^+(aq) + OH^-(aq)} \;\;\; \Delta H<0\] \[ \chem{NaHCO_3(s)\rightarrow Na^+(aq) + HCO_3^-(aq)} \;\;\; \Delta H>0 \]

Spontaneity Conditions for a Process At Constant \(T\) and \(P\)

\[ \Delta G = \Delta H - T \Delta S \lt 0 \]

Helmholtz Function (or Helmholtz Free Energy)

• Similarly to the Gibbs' Function we can define the Helmholtz Function at constant \(P\) and \(T\) \[ A \equiv U-TS \] \[ \Delta A = \Delta U - T\Delta S\] \[\text{For spontaneous reactions: }\Delta A = \Delta U - T\Delta S \lt 0 \]

Example 6.1

Given the following data at \(298K\), calculate \(\Delta G^\circ\) at \(298K\) for the following reaction: \[ \chem{C_2H_4(g) + H_2(g) \rightarrow C_2H_6(g)} \]

Combining the First and Second Laws - Maxwell's Relations

• For a reversible change \(dS=\frac{dq}{T}\) and therefore \[ dq=TdS \]
• Since \(dw=-PdV\) for a reversible expansion, then since \(dU=dq+dw\) \[ dU=TdS-PdV \]

Maxwell Relations

• Because \(dU=TdS-PdV\) we know that internal energy has natural variables of \(S\) and \(V\) so we can calculate the total differential \[ dU=\left(\frac{\partial U}{\partial S}\right)_VdS+\left(\frac{\partial U}{\partial V}\right)_SdV \]
• By inspection of the two equations above we can see that \[ \left(\frac{\partial U}{\partial S}\right)_V=T \text{ and } \left(\frac{\partial U}{\partial V}\right)_S=-P \]

First Maxwell Relation

• Internal energy is a state function so it's total differential obeys Euler's Relation (remember that?) \[ \text{Euler's Relation: } \left\{ \begin{align} df(x,y) = P\,dx + Q\,dy \\ \left(\frac{\partial P}{\partial y}\right)_x = \left(\frac{\partial Q}{\partial x}\right)_y \end{align} \right. \] \[ \left[ \frac{\partial}{\partial V}\left(\frac{\partial U}{\partial S}\right)_V\right]_S=\left[\frac{\partial}{\partial S}\left(\frac{\partial U}{\partial V}\right)_S\right]_V \]
• Substituting in what we know from the last slide \[ \left[ \frac{\partial}{\partial V}\left(T\right)_V \right]_S = \left[ \frac{\partial}{\partial S} \left(-P\right)\right]_V \] \[ \left(\frac{\partial T}{\partial V}\right)_S = -\left(\frac{\partial P}{\partial S}\right)_V \]

Building Another Maxwell Relation

\[ H\equiv U+PV \] \[ dH=dU+PdV+VdP \] \[ dH = TdS-PdV+PdV+VdP = TdS+VdP\] \[ dH=\left(\frac{\partial H}{\partial S}\right)_PdS+\left( \frac{\partial H}{\partial P}\right)_SdV \] \[ \left(\frac{\partial H}{\partial S}\right)_P=T \text{ and } \left(\frac{\partial H}{\partial P}\right)_S=V \] Note that: \[ \left(\frac{\partial U}{\partial S}\right)_V=T=\left(\frac{\partial H}{\partial S}\right)_P \]

Our Second Maxwell Relation

\[ \left[ \frac{\partial}{\partial P}\left( \frac{\partial H}{\partial S}\right)_P\right]_S = \left[ \frac{\partial}{\partial S}\left( \frac{\partial H}{\partial P}\right)_S\right]_P \] \[ \left(\frac{\partial T}{\partial P}\right)_S = \left(\frac{\partial V}{\partial S}\right)_P \]

Four Maxwell Relations

Example 6.2

Show that \[ \left(\frac{\partial U}{\partial V}\right)_T=T\frac{\alpha}{\kappa_T}-P \]

\(\Delta A\), \(\Delta G\), and Maximum Work

• \(A\) and \(G\) are often referred to as free energy functions.
• This is because \(\Delta A\) measures the maximum work and \(\Delta G\) measures the non P-V work. Let's show it...
• Let's start with \(A\): \( dA=dU-TdS-SdT \)
• Substituting the combined first and second laws for \(dU\) we find that \[ dA=TdS+dw-TdS-SdT = dw-SdT \] or at constant temperature \(dA=dw\)
• The only assumption here is that the change is reversible and \(dw\) for a reversible change is the maximum amount of work, therefore \(dA\) gives the maximum work at constant pressure

What About \(G\)

\[ G=H-TS=U+PV-TS \] \[ dG=dU+PdV+VdP-TdS-SdT \]

• Using \(dU=dq+dw\) where \(dq=TdS\) and \(dw\) is split into two terms \(dw_{PV}\) and \(dw_e\) \[ dU=TdS+dw_{PV}+dw_e \]
• So\(\require{cancel}\) \[ dG=\cancel{TdS}\cancel{-PdV}+dw_e+\cancel{PdV}+VdP\cancel{-TdS}-SdT \] \[ dG=+dw_e+VdP-SdT \]
• So, at constant temperature and pressure, \(dG=dw_e\), which is the maximum amount of work at constant temperature and pressure.

Volume Dependence of Helmholtz Energy

• How do we determine the changes in the Helmholtz function as the volume changes at constant pressure? \[ \Delta A = \int_{V_1}^{V_2} \left(\frac{\partial A}{\partial V}\right)_TdV \]
• How do we solve the differential? Let's start with \(A=U-TS\) \[ dA=dU-TdS-SdT \]

Working on our Derivation

• Now we can use \(dU=TdS-PdV\) which assumes a reversible change and only PV work \[ dA=\cancel{TdS}-PdV-\cancel{TdS}-SdT =-PdV-SdT\]
• So, \(A\) has natural variables of \(V\) and \(T\) so \[ dA=\left(\frac{\partial A}{\partial V}\right)_TdV + \left(\frac{\partial A}{\partial T}\right)_VdT \]
• This means that \[ \left(\frac{\partial A}{\partial V}\right)_T=-P \text{ and } \left(\frac{\partial A}{\partial T}\right)_V=-S \]

Completing our Derivation

• Plugging what we just learned into our integral \[ \Delta A = - \int_{V_1}^{V_2} P\,dV \]
• If pressure is independent of temperature it can be pulled out of the integral \[ \Delta A = -P\int_{V_1}^{V_2} dV=-P\left(V_2-V_1\right) \]

Further Extension

• From here we can get our third Maxwell relation \[ \left(\frac{\partial P}{\partial T}\right)_V = \left(\frac{\partial S}{\partial V}\right)_T \] and \[ \left(\frac{\partial P}{\partial T}\right)_V = \frac{\alpha}{\kappa_T} \]

Example 6.3

Calculate \(\Delta A\) for the isothermal expansion of 1.00 mol of an ideal gas from 10.0 L to 25.0 L at 298 K.

Pressure Dependence of Gibbs' Energy

• To find the pressure and temperature dependence of \(G\) let's start at \(G=U+PV-TS\) \[ dG=dU+PdV+VdP-TdS-SdT \]
• Substituting in \(dU=TdS-PdV\) \[ dG=\cancel{TdS}-\cancel{PdV}+\cancel{PdV}+VdP-\cancel{TdS}-SdT = VdP-SdT \]
• This leads to a total differential \[ dG = \left(\frac{\partial G}{\partial P}\right)_TdP+\left(\frac{\partial G}{\partial T}\right)_PdT \]

Continuing the Derivation

• From the differential expressions \[ \left(\frac{\partial G}{\partial P}\right)_T=V \text{ and } \left(\frac{\partial G}{\partial T}\right)_P=-S \]
• The fourth Maxwell Relation is \[ \left(\frac{\partial V}{\partial T}\right)_P = \left(\frac{\partial S}{\partial P}\right)_T \] and therefore \[ \left(\frac{\partial V}{\partial T}\right)_P=V\alpha \]

Result

• The pressure dependence of \(G\) is given by the pressure derivative at constant temperature \[ \left(\frac{\partial G}{\partial P}\right)_T=V \] which is simply the molar volume
• For fairly incompressible substances (liquids and solids) the molar volume will be essentiall constant over a moderate pressure range

Example 6.4

Temperature Dependence of \(A\) and \(G\)

• In differential form we already know that \[ dA=-PdV-SdT \text{ and } dG=VdP-SdT \]
• So, by inspection we know that \[ \left(\frac{\partial A}{\partial T}\right)_V=-S \text{ and } \left(\frac{\partial G}{\partial T}\right)_P=-S \]

Problem

• Based on the previous results \[ \Delta A = - \int_{T_1}^{T_2}\left(\frac{\partial A}{\partial T}\right)_VdT=-\int_{T_1}^{T_2}S\,dT \] and \[ \Delta G = - \int_{T_1}^{T_2}\left(\frac{\partial G}{\partial T}\right)_PdT=-\int_{T_1}^{T_2}S\,dT \]
• The problem is that the temperature dependence of entropy would need to be known...
• Let's try something else

New Method of Getting There

• Let's start with \[ A=U-TS \;\;\text{ and }\;\; G=H-TS \]
• Dividing by \(T\) gives \[ \frac{A}{T}=\frac{U}{T}-S \;\;\text{ and }\;\; \frac{G}{T}=\frac{H}{T}-S \]
• Now we can differentiate with respect to \(T\) at constant \(V\) \[ \left( \frac{\partial \left(\frac{A}{T}\right)}{\partial T}\right)_V = -\frac{U}{T^2} \;\; \text{ and } \;\; \left(\frac{\partial \left(\frac{G}{T}\right)}{\partial T}\right)_P = -\frac{H}{T^2} \]

Simplification

\[ \left( \frac{\partial \left(\frac{A}{T}\right)}{\partial T}\right)_V = -\frac{U}{T^2} \;\; \text{ and } \;\; \left(\frac{\partial \left(\frac{G}{T}\right)}{\partial T}\right)_P = -\frac{H}{T^2} \]

• Let's try instead by differentiating with respect to \(\bfrac{1}{T}\) \[ \left( \frac{\partial \left(\frac{A}{T}\right)}{\partial \left(\frac{1}{T}\right)}\right)_V = U \;\; \text{ and } \;\; \left(\frac{\partial \left(\frac{G}{T}\right)}{\partial \left(\frac{1}{T}\right)}\right)_P = H \]
• From this second expression \[ d\left(\frac{G}{T}\right) = H\, d\left(\frac{1}{T}\right) \;\;\text{ or }\;\; d\left(\frac{\Delta G}{T}\right) = \Delta H\, d\left(\frac{1}{T}\right) \]

Solving the Integral

• We can now integrate this expression \[ \int_{T_1}^{T_2} d\left(\frac{\Delta G}{T}\right) = \Delta H \int_{T_1}^{T_2} d\left(\frac{1}{T}\right) \] \[ \frac{\Delta G_{T_2}}{T_2} - \frac{\Delta G_{T_1}}{T_1} = \Delta H \left(\frac{1}{T_2}-\frac{1}{T_1}\right) \] This is the Gibbs-Helmholtz Equation
• We can derive an equivalent equation for the Helmholtz function \[ \frac{\Delta A_{T_2}}{T_2} - \frac{\Delta A_{T_1}}{T_1} = \Delta H \left(\frac{1}{T_2}-\frac{1}{T_1}\right) \]

Example 6.5

Given the following data at 298 K, calculate \(\Delta G\) at 500 K for the following reaction: \[ \chem{CH_4(g)+2O_2(g)\rightarrow CO_2(g)+H_2O(g)} \]

When Two Variables Change at Once

• We have only considered what happens to a function when one variable changes and others are constant
• In the real world, things are rarely this simple
• Fortunately for state variables, we can considered the changes independently and add the results together \[ dG=\left(\frac{\partial G}{\partial P}\right)_TdP + \left(\frac{\partial G}{\partial T}\right)_PdT \]
• Since \(G\) is a state variable, the pathway linking the initial and final states is unimportant.

Example 6.6

Deriving an Expression for a Partial Derivative (Type III)

• Let's look at a situation in which we have two function of the same two variables, such as \(z(x,y)\) and \(w(x,y)\)
• What happens to \(z\) if \(w\) is held constant, but \(x\) is changed?
• Let's start with our total differential \[ dz = \left(\frac{\partial z}{\partial x} \right)_ydx + \left(\frac{\partial z}{\partial y}\right)_xdy \]
• \(z\) can also be considered a function of \(x\) and \(w(x,y)\) so that \[ dz = \left(\frac{\partial z}{\partial x} \right)_wdx + \left(\frac{\partial z}{\partial w}\right)_xdw \]

Continuing the Derivation

• Both of these total derivatives equal \(dz\) so they must equal each other \[ \left(\frac{\partial z}{\partial x} \right)_ydx + \left(\frac{\partial z}{\partial y}\right)_xdy = \left(\frac{\partial z}{\partial x} \right)_wdx + \left(\frac{\partial z}{\partial w}\right)_xdw \]
• If, as we said, we constrain the system such that \(w\) is constant (\(dw=0\)) \[ \left(\frac{\partial z}{\partial x} \right)_ydx + \left(\frac{\partial z}{\partial y}\right)_xdy = \left(\frac{\partial z}{\partial x} \right)_wdx \]

More Derivation

• Now, since \(w\) is a function of \(x\) and \(y\) \[ dw = \left(\frac{\partial w}{\partial x}\right)_y dx + \left(\frac{\partial w}{\partial y}\right)_x dy \]
• Since \(w\) is being held constant \[ 0 = \left(\frac{\partial w}{\partial x}\right)_y dx + \left(\frac{\partial w}{\partial y}\right)_x dy \]
• Rearranging and solving for \(dy\) we find that \[ \left(\frac{\partial w}{\partial y}\right)_x dy = - \left(\frac{\partial w}{\partial x}\right)_y dx \] \[ dy = - \left(\frac{\partial w}{\partial x}\right)_y \left(\frac{\partial y}{\partial w}\right)_x dx \]

Still More Derivation

• Plugging our \(dy\) expression into our earlier expression \[ \left(\frac{\partial z}{\partial x} \right)_ydx + \left(\frac{\partial z}{\partial y}\right)_x \left[ - \left(\frac{\partial w}{\partial x}\right)_y \left(\frac{\partial y}{\partial w}\right)_x dx \right] = \left(\frac{\partial z}{\partial x} \right)_wdx \] \[ \left(\frac{\partial z}{\partial x} \right)_ydx - \left(\frac{\partial z}{\partial y}\right)_x \left(\frac{\partial y}{\partial w}\right)_x \left(\frac{\partial w}{\partial x}\right)_y dx = \left(\frac{\partial z}{\partial x} \right)_wdx \] \[ \left(\frac{\partial z}{\partial x} \right)_ydx - \left(\frac{\partial z}{\partial w}\right)_x \left(\frac{\partial w}{\partial x}\right)_y dx = \left(\frac{\partial z}{\partial x} \right)_wdx \]
• If \(dx \ne 0\) then this implies that \[ \left(\frac{\partial z}{\partial x} \right)_y - \left(\frac{\partial z}{\partial w}\right)_x \left(\frac{\partial w}{\partial x}\right)_y = \left(\frac{\partial z}{\partial x} \right)_w \]

Finishing the Derivation

• Our previous result can be rewritten \[ \left(\frac{\partial z}{\partial x} \right)_y = \left(\frac{\partial z}{\partial x} \right)_w + \left(\frac{\partial z}{\partial w}\right)_x \left(\frac{\partial w}{\partial x}\right)_y \]

The Difference Between \(C_P\) and \(C_V\)

• Let's look at \(C_P-C_V\)
• Let's start with the definitions \[ C_P=\left(\frac{\partial H}{\partial T}\right)_P \;\;\text{ and }\;\; C_V=\left(\frac{\partial U}{\partial T}\right)_V \]
• This means that the difference is \[ C_P-C_V = \left(\frac{\partial H}{\partial T}\right)_P - \left(\frac{\partial U}{\partial T}\right)_V \]

Simplifying the Expression

• We can use the defintion of \(H\) to simplify this expression \[ H=U+PV \Rightarrow dH=dU+PdV+VdP \]
• Deviding this expression by \(dT\) and holding \(P\) constant gives \[ \left. \frac{dH}{dT}\right|_P = \left. \frac{dU}{dT}\right|_P + P\left. \frac{dV}{dT}\right|_P + V\left. \frac{dP}{dT}\right|_P \] \[ \left( \frac{\partial H}{\partial T}\right)_P = \left( \frac{\partial U}{\partial T}\right)_P + P\left( \frac{\partial V}{\partial T}\right)_P \]
• From previous section we know that \[ \left(\frac{\partial H}{\partial T}\right)_P=C_P \;\;\text{ and }\;\; \left(\frac{\partial V}{\partial T}\right)_P=V\alpha \]

Further Simplification to Our Equation

• Substituting these expression into our differential equation gives \[ C_P = \left(\frac{\partial U}{\partial T}\right)_P + PV\alpha \]
• In order to evaluate the partial derivative in this equation we need to start with the total differential of \(U\) \[ dU=\left(\frac{\partial U}{\partial V}\right)_TdV - \left(\frac{\partial U}{\partial T}\right)_V dT \]
• Dividing by \(dT\) and holding \(P\) constant we get \[ \left. \frac{dU}{dT} \right|_P = \left(\frac{\partial U}{\partial V}\right)_T \left. \frac{dV}{dT} \right|_P + \left( \frac{\partial U}{\partial T}\right)_V \left. \frac{dT}{dT} \right|_P \]

Getting Closer to Our Solution

• Simplifying our expression we find that \[ \left( \frac{\partial U}{\partial T} \right)_P = \left(\frac{\partial U}{\partial V}\right)_T \left( \frac{\partial V}{\partial T} \right)_P + \left( \frac{\partial U}{\partial T}\right)_V \]
• Remember from previous section that \[ \left(\frac{\partial U}{\partial T}\right)_V=C_V \;\;\text{ and }\;\; \left(\frac{\partial V}{\partial T}\right)_P=V\alpha \]
• So, plugging these in we find that \[ \left( \frac{\partial U}{\partial T} \right)_P = \left(\frac{\partial U}{\partial V}\right)_T V\alpha + C_V \]

We need to derive another partial derivative

• Now we need to find a expression for \( \left(\frac{\partial U}{\partial V}\right)_T \)
• Let's start with \(dU=TdS-PdV\) and divide both sides by \(dV\) at constant \(T\) \[ \left. \frac{dU}{dV} \right|_T = T \left. \frac{dS}{dV} \right|_T - P \left. \frac{dV}{dV} \right|_T \] \[ \left( \frac{\partial U}{\partial V} \right)_T = T \left( \frac{\partial S}{\partial V} \right)_T - P \]

Simplification with a Maxwell Relation

\( \left( \frac{\partial U}{\partial V} \right)_T = T \left( \frac{\partial S}{\partial V} \right)_T - P \)

• Let's use the following Maxwell relation \[ \left(\frac{\partial S}{\partial V}\right)_T = \left( \frac{\partial P}{\partial T}\right)_V \]
• Substituting this in we find that \[ \left( \frac{\partial U}{\partial V} \right)_T = T \left( \frac{\partial P}{\partial T} \right)_V - P \]
• Remember that \( \left(\frac{\partial P}{\partial T}\right)_V = \frac{\alpha}{\kappa_T} \) so that \[ \left( \frac{\partial U}{\partial V} \right)_T = T \left( \frac{\alpha}{\kappa_T} \right) - P \]

Plugging stuff in now

• Plugging our last result into our result for \( \left( \frac{\partial U}{\partial T}\right)_P \) we find that \[ \left( \frac{\partial U}{\partial T} \right)_P = \left(\frac{\partial U}{\partial V}\right)_T V\alpha + C_V \] \[ \left( \frac{\partial U}{\partial T} \right)_P = \left[ T\frac{\alpha}{\kappa_T} -P \right] V\alpha + C_V \] \[ \left( \frac{\partial U}{\partial T} \right)_P = \frac{T V \alpha^2}{\kappa_T} -PV\alpha + C_V \]
• Plugging this back into our \(C_P\) equation \[ C_P = \left[ \frac{TV\alpha^2}{\kappa_T} - \cancel{PV\alpha} + C_V \right] + \cancel{PV\alpha} = \frac{TV\alpha^2}{\kappa_T} + C_V \]

Almost Done

\( C_P = \frac{TV\alpha^2}{\kappa_T} + C_V \)

• Subtracting \(C_V\) to the other side we find that \[ C_P - C_V = \frac{TV\alpha^2}{\kappa_T} \]
• Done!!

Evaluating Our Expression for an Ideal Gas

\( C_P - C_V = \frac{TV\alpha^2}{\kappa_T} \)

• Let's evaluate our expression for an ideal gas
• For an ideal gas \[ \alpha=\frac{1}{T} \;\;\text{ and }\;\; \kappa_T=\frac{1}{P} \]
• So \[ C_P - C_V = \frac{TV\left(\frac{1}{T}\right)^2}{\left(\frac{1}{P}\right)} \] \[ C_P - C_V = \frac{PVT}{T^2} = \frac{PV}{T} = R \]